4-1.Complex numbers
hard

Let $\alpha=8-14 i , A=\left\{ z \in C : \frac{\alpha z -\bar{\alpha} \overline{ z }}{ z ^2-(\overline{ z })^2-112 i }=1\right\}$ and $B =\{ z \in C 😐 z +3 i |=4\}$ Then $\sum_{z \in A \cap B}(\operatorname{Re} z-\operatorname{Im} z)$ is equal to $...............$.

A

$14$

B

$13$

C

$12$

D

$11$

(JEE MAIN-2023)

Solution

$\alpha=8-14 i$

$z=x+i y$

$a z=(8 x+14 y)+i(-14 x+8 y)$

$z +\overline{ z }=2 x \quad z -\overline{ z }=2 iy$

Set A: $\frac{2 i(-14 x+8 y)}{i(4 x y-112)}=1$

$(x-4)(y+7)=0$

$x=4 \quad \text { or } \quad y=-7$

Set B: $x^2+(y+3)^2=16$

when $x=4 \quad y=-3$

when $y =-7 \quad x =0$

$\therefore A \cap B=\{4-3 i, 0-7 i\}$

So, $\sum_{z \in A \cap B}(\operatorname{Re} z-\operatorname{Im} z)=4-(-3)+(0-(-7))=14$

Standard 11
Mathematics

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